$f(x)=\dfrac{2-x}{x+1}$ $g(x)=18-3x$ Write $(f\circ g)(x)$ as an expression in terms of $x$. $(f\circ g)(x)=$
Explanation: First, let's write $(f\circ g)(x)$ as $f(g(x))$. Next, we write $g(x)$ as the input to function $f$. $f({g(x)})=\dfrac{2-{g(x)}}{{g(x)}+1}$ Since $g(x)=18-3x$, this becomes: f ( g ( x ) ) = 2 − ( 18 − 3 x ) ( 18 − 3 x ) + 1 = 2 − 18 + 3 x 19 − 3 x = 3 x − 16 19 − 3 x \begin{aligned} f({g(x)})&=\dfrac{2-({18-3x})}{({18-3x})+1}\\ &=\dfrac{2-18+3x}{19-3x}\\ \\ &=\dfrac{3x-16}{19-3x}\ \\ \end{aligned} Note: We simplified the result to obtain a nicer expression, but this is not necessary. The answer: $(f \circ g)(x)=\dfrac{3x-16}{19-3x}$